230 Kth Smallest Element in a BST

Author: int32bit

README.md

@@ -126,6 +126,7 @@
 + [219 Contains Duplicate II(hashmap)](algorithms/ContainsDuplicateII)
 + [226 Invert Binary Tree(递归，树)](algorithms/InvertBinaryTree)
 + [228 Summary Ranges](algorithms/SummaryRanges)
++ [230 Kth Smallest Element in a BST(BST、中序遍历)](algorithms/KthSmallestElementinaBST)
 + [231 Power of Two(位运算，二进制1的个数)](algorithms/PowerofTwo)
 
 ## Database

algorithms/KthSmallestElementinaBST/README.md

@@ -0,0 +1,63 @@
+## Kth Smallest Element in a BST
+
+Given a binary search tree, write a function `kthSmallest` to find the kth smallest element in it.
+
+Note: 
+You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
+
+Follow up:
+What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
+
+Hint:
+
+* Try to utilize the property of a BST.
+* What if you could modify the BST node's structure?
+* The optimal runtime complexity is O(height of BST).
+
+Credits:
+Special thanks to @ts for adding this problem and creating all test cases.
+
+## Solution
+
+中序遍历，更新k值，访问时`k--`， 若此时`k == 0`, 则当前值即为结果
+
+```cpp
+public:
+	int kthSmallest(TreeNode *root, int k) {
+		int result;
+		/*
+		if (root == nullptr) {
+			return 0;
+		}
+		*/
+		kthSmallest(root, k, result);
+		return result;
+	}
+private:
+	/* k为需要查找的第k小的数， result为结果，返回true表示已经查找到
+	 * 注意k为引用类型，否则k的值更新不能反馈到上一栈调用
+	*/
+	bool kthSmallest(TreeNode *root, int &k, int &result) {
+		if (root->left) {
+			if (kthSmallest(root->left, k, result)) {
+				return true;
+			}
+		}
+		k -= 1;
+		if (k == 0) {
+			result = root->val;
+			return true;
+		}
+		if (root->right) {
+			if (kthSmallest(root->right, k, result)) {
+				return true;
+			}
+		}
+		return false;
+	}
+};
+```
+
+## 优化
+
+如果允许修改树结构，保存节点左孩子个数，是否可以优化呢？

algorithms/KthSmallestElementinaBST/tree.cpp

@@ -0,0 +1,72 @@
+#include <iostream>
+#include <vector>
+#include <string>
+#include <algorithm>
+#include <cstdio>
+using namespace std;
+struct TreeNode {
+	int val;
+	TreeNode *left;
+	TreeNode *right;
+	TreeNode(int x) : val(x), left(nullptr), right(nullptr){}
+};
+class Solution {
+public:
+	int kthSmallest(TreeNode *root, int k) {
+		int result;
+		/*
+		if (root == nullptr) {
+			return 0;
+		}
+		*/
+		kthSmallest(root, k, result);
+		return result;
+	}
+private:
+	bool kthSmallest(TreeNode *root, int &k, int &result) {
+		if (root->left) {
+			if (kthSmallest(root->left, k, result)) {
+				return true;
+			}
+		}
+		k -= 1;
+		if (k == 0) {
+			result = root->val;
+			return true;
+		}
+		if (root->right) {
+			if (kthSmallest(root->right, k, result)) {
+				return true;
+			}
+		}
+		return false;
+	}
+};
+TreeNode *mk_node(int val)
+{
+	return new TreeNode(val);
+}
+TreeNode *mk_child(TreeNode *root, TreeNode *left, TreeNode *right)
+{
+	root->left = left;
+	root->right = right;
+	return root;
+}
+TreeNode *mk_child(TreeNode *root, int left, int right)
+{
+	return mk_child(root, new TreeNode(left), new TreeNode(right));
+}
+TreeNode *mk_child(int root, int left, int right)
+{
+	return mk_child(new TreeNode(root), new TreeNode(left), new TreeNode(right));
+}
+int main(int argc, char **argv)
+{
+	Solution solution;
+	TreeNode *root = mk_child(6, 4, 8);
+	mk_child(root->left, 3, 5);
+	mk_child(root->right, 7, 9);
+	for (int i = 1; i <= 7; ++i)
+		printf("%d\n", solution.kthSmallest(root, i));
+	return 0;
+}